Effect of pressure at ocean-bottom

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supratim1
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Main Question or Discussion Point

If we take a steel container, fill it completely with water and seal it back (all this at sea-level), and then take it to bottom of sea, what will happen to the container? Will it be crushed? What equation should we use to relate initial and final volume? Kindly help.
 

Answers and Replies

  • #2
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If water would be perfectly incompressible, nothing would happen. As water can be compressed a bit, you will get some stress at the steel container. The details depend on the container structrue. Most probable things:
- it will deform a bit and decrease its volume. If it is too weak, the sealing might break and you get an open container
- it resists and does not deform in a significant way
 
  • #3
supratim1
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well thanks. my doubt is cleared now.
 
  • #4
russ_watters
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You could check to see how water and steel differ in compressibility...
 
  • #5
supratim1
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yes sir, surely i will. thank you.
 
  • #6
K^2
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Compresssibility of steel is irrelevant. Strength of the barrel as well. The pressures are just too high for it to make any difference.

The only value that's relevant is bulk modulus of Water, which is K=2.2E9 Pa. Bulk modulus is defined with following equation.

[tex]K=-V\frac{dP}{dV}[/tex]

If you solve this as diff. eq. and substitute in 11km as depth, you'll get 5% change in volume of the barrel due to pressure. So barrel will deform, but not a whole lot.
 
  • #7
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For steel, the value is larger by two orders of magnitude. There are submarines which can reach the bottom of the mariana trench with humans inside, so the pressure inside and the compression of the steel is negligible.
 
  • #8
K^2
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You are not interested in bulk modulus for steel for that. You need to figure out what the stress is going to be on structure overall. It's a far more complicated problem.

I assumed that the container in the problem is far bellow its crush depth, because otherwise, the question makes no sense. The bulk modulus of container material is irrelevant either way.
 
  • #9
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If the material is strong enough to stay in its initial shape, you can use the bulk modulus to check if there are any significant effects. The result is "no", which is what I stated in my post.

I assumed that the container in the problem is far bellow its crush depth
As stated earlier, it depends on the container structure.
 
  • #10
K^2
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Material's bulk modulus need not relate to it's young's and shear moduli. A barrel can deform quite a bit without being irreversibly crushed or changing volume of actual material. Any way you twist it, bulk modulus of the container material is absolutely irrelevant. Only bulk modulus of the content fluid makes a difference.
 
  • #11
D H
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Only bulk modulus of the content fluid makes a difference.
Nonsense.

Submarines have gone to the bottom of the Mariana Trench (pressure: about 108 Pa) with the content fluid being air (bulk modulus: about 105 Pa). The sub would be completely crushed if what you said was true. The sub isn't crushed. The near incompressibility of the submarine's massive hull keeps that cabin air at one atmosphere throughout the trip. This eliminates the need for decompression stops on the way up.
 
  • #12
K^2
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Nonsense.

Submarines have gone to the bottom of the Mariana Trench (pressure: about 108 Pa) with the content fluid being air (bulk modulus: about 105 Pa). The sub would be completely crushed if what you said was true. The sub isn't crushed. The near incompressibility of the submarine's massive hull keeps that cabin air at one atmosphere throughout the trip. This eliminates the need for decompression stops on the way up.
Way to quote something out of context without reading the whole exchange.

Does bulk modulus of the steel the sub is made out of make any difference? No. The only bulk modulus that's relevant is that of contents.
 

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